Question

7.00g of Compound X with molecular formula C5H10 are burned in a constant-pressure calorimeter containing 35.00kg of water at 25°C. The temperature of the water is observed to rise by 2.113°C. (You may assume all the heat released by the reaction is absorbed by the water, and none by the calorimeter itself.) Calculate the standard heat of formation of Compound X at 25°C. Be sure your answer has a unit symbol, if necessary, and round it to the correct number of significant digits.

Answer 1

Explanation:

The given data is as follows.

      mass of water = 35.00 kg = 35 × 1000 g = 35000 g

      specific heat of water = 4.186 $J/g ^{0}C$

      change in temperature = 2.113 $^{0}C$

Formula to calculate heat change is as follows.

                       q = $m \times C \times \Delta T$

Putting the given values into the above formula as follows.

                        q = $m \times C \times \Delta T$

                        q = $35000 g \times 4.186 J/g ^{0}C \times 2.113^{0}C$

                           = 309131.9 J

or,                        = $\frac{309131.9 J \times 1 kg}{1000 J}$

                            = 309.1 kJ

As 7.00 g of compound 'X' is giving 309.131 kJ of heat. Therefore, 1 mole of $C_{5}H_{10}$ (molar mass = 70 g) will give the heat as follows.

                      $\frac{309.1}{7} \times 70 kJ/mol$

                    = 3091 kJ/mol

So, the reaction equation will be as follows.

      $C_{5}H_{10}(g) + \frac{15}{2}O_{2} \rightarrow 5CO_{2}(g) + 5H_{2}O(g)$    

Value of $\Delta H_{combustion}$ = 3091 kJ/mol for the reaction.

Standard enthalpy values for formation of included atoms or molecules into this reaction are as follows.

       $CO_{2}(g)$ = -393.5 kJ/mol,      $O_{2}(g)$ = 0 kJ/mol

       $H_{2}O(g)$ = -241.8 kJ/mol

Formula to calculate standard heat of formation of compound $C_{5}H_{10}$ is as follows.

          $\Delta H_C_{5}H_{10} = E\Delta H^{0}_f(products) - E\Delta H^{0}_f(reactants)$

                        = $[5 times \Delta H^{0}_f{CO_{2}} + 5 times \Delta H^{0}_f{H_{2}O}] - [5 times \Delta H^{0}_f{C_{5}H_{10}} + \frac{15}{2} \Delta H^{0}_f{O_{2}}$

              3091 kJ/mol = $[(5 times -393.5 kJ/mol) + (5 \times -241.8 kJ/mol)] - [5 times \Delta H^{0}_f{C_{5}H_{10}} + (\frac{15}{2} \times 0)]$

             $-\Delta H^{0}_f{C_{5}H_{10}}$ = -3091 kJ/mol + 1967.5 kJ/mol + 1209 kJ/mol

                                             = 85.5 kJ/mol

or,             $\Delta H^{0}_f{C_{5}H_{10}}$ = -85.5 kJ/mol

Thus, we can conclude that the standard heat of formation of given compound X is -85.5 kJ/mol.

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